1. **State the problem:** Find the vector equation of the line given the symmetric equations: $$\frac{x - 2}{3} = \frac{y - 2}{2} = \frac{z + 1}{4}$$ 2. **Recall the formula:** The symmetric form of a line is: $$\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$$ where $(x_0, y_0, z_0)$ is a point on the line and $(a, b, c)$ is the direction vector. 3. **Identify point and direction vector:** From the given equation: - Point on the line: $P_0 = (2, 2, -1)$ - Direction vector: $\vec{d} = \langle 3, 2, 4 \rangle$ 4. **Write the vector equation:** The vector equation of the line is: $$\vec{r} = \vec{r_0} + t\vec{d}$$ where $\vec{r_0} = 2\vec{i} + 2\vec{j} - \vec{k}$ and $\vec{d} = 3\vec{i} + 2\vec{j} + 4\vec{k}$. So, $$\vec{r} = 2\vec{i} + 2\vec{j} - \vec{k} + t(3\vec{i} + 2\vec{j} + 4\vec{k})$$ 5. **Compare with options:** Option C matches exactly. **Final answer:** Option C $$\boxed{\vec{v} = 2\vec{i} + 2\vec{j} - \vec{k} + t(3\vec{i} + 2\vec{j} + 4\vec{k})}$$